By the fundamental theorem of calculus (see, e.g., [1]}), it follows that
\(\int _{\prod \limits _{j=1}^{m} [0, x_j)} \, d\nu _{\mathbf {1}_m}(t) = \int _{\prod \limits _{j=1}^{m} [0, x_j)} \, d\mu _{\mathbf {1}_m}(t)\) 
